Question

Consider the following table which shows different baskets of tennis balls:

Baskets	Number of golf balls (Population)
1	10
2	18
3	7
4	11
5	6

(a) List all samples of size 2, and compute the mean of each sample.

(b) Compute the mean of the distribution of the sample mean and the population mean. Compare the two values.

(c) Compare the dispersion in the population with that of the sample mean.

Answer 1

(a)

All samples of size 2 are,

(10, 18) , (10, 7) (10, 11) (10, 6)

(18, 7) , (18, 11) , (18, 6)

(7, 11) , (7, 6)

(11, 6)

The mean of all the samples are

(10 + 18)/2 , (10 + 7)/2 , (10 + 11)/2, (10 + 6)/2, (18 + 7)/2, (18 + 11)/2, (18 + 6)/2, (7+11)/2, (7 + 6)/2, (11 + 6) / 2

= 14, 8.5, 10.5, 8, 12.5, 14.5, 12, 9, 6.5, 8.5

(b)

The mean of the distribution of the sample mean = (14 + 8.5 + 10.5 + 8 + 12.5 +14.5 + 12 +9 +6.5 +8.5) / 10 = 10.4

Population mean = (10 + 18 + 7 + 11 + 6) / 5 = 10.4

Thus, the mean of the distribution of the sample mean is equal to the population mean.

(c)

The variance of the distribution of the sample mean = [(14 - 10.4)² + (8.5 - 10.4)² + (10.5 - 10.4)² + (8 - 10.4)² + (12.5 - 10.4)² + (14.5 - 10.4)² + (12 - 10.4)² + (9 - 10.4)² + (6.5 - 10.4)² + (8.5 - 10.4)² ] / 10 = 6.69

Population variance = [(10 - 10.4)² + (18 - 10.4)² + (7 - 10.4)² + (11 - 10.4)² + (6 - 10.4)² ] / 5 = 17.84

Thus, the dispersion of the distribution of the sample mean is less than that of the population mean.